So for this, so let's quickly do these questions. We have, we count the number of votes we've already got, right, sum up all these numbers, and we got 40, 45, I think. We got 45, and that's 90% of the votes, so we know that there are five more votes to go. And five more votes to go, Alex can definitely get all those five and win. Same for Bella, she got to 16, and Calvin can got to 15, more than Alex. So three of them have, still have the chance of winning the elections, and so it's C. Okay, good. Let's go to the next one. Oh, I mean, questions, not the next questions. Okay. So last week we talk about the basic of algebra, and today we delve more into problem solving. So what's the step of solving a problem algebraically? First, definitely read the question carefully. Then the next step is to identify the unknowns, and choose a variable wisely for the unknowns, right? This is something that we emphasized last week. There could be questions, there are a lot of unknowns, but we do not need every variable for each unknown. And the way, and there are more than one way to assign variables, and if we choose it wisely, we're gonna save a lot on computation time later. And then set up the equations. Lots of time we have to deal with a system of equations, as a clear distinction between linear versus non-linear, one variables versus multi-variables, and we talk about the technique to solve the equation last time, and we solve them, define the variables, and double-check the answer. Sorry. Sorry, there's something wrong with this thing. Sorry. So common types of algebra problems that you may find in Maths Kangaroo. First type is finding missing numbers on the diagrams. Then you can use a variable for each unknown, or a minimum number of variables, and express other unknowns in terms of these variables. We also, a lot of system, we have to find numbers with certain properties, certain constraints. There are different ways to solve that, but many times there's some things that are pretty useful. It's called, we go through some problems in this lesson that involve using generalized forms. And then this is something that you've probably seen earlier in seventh and eighth grade, percents, ratio, proportion. These are very common problems in photographs for word problems, right? Percent increase, percent decrease, multi-ratios, proportion, you can expect to see this. Then there's another type of questions that you see a lot in level nine and 10 is velocity. So velocity, it's pretty much like proportion and rate, right, because velocity is the rate of change of distance over time. But the interesting about velocity is once we add direction to it, right? If it's a scalar quantity, then we call it a speed, but if we add a direction to it, we call it velocity. It's just like a vector. It has a magnitude and direction. We could have a perpendicular direction. We could have something moving in a circular path. So in question like that, it's helpful to draw a diagram. And also sometimes we have to work with relative speed. If two objects move in the same directions, right, relative speed is the difference between the two speeds. And when they move in opposite directions, then relative speed is the sum of the two speeds, right? If you study physics, you know that it's a frame of reference. So we definitely are gonna have a few interesting questions about velocity in this. And this is like by no means everything about algebra, right, when we go to number theory, we go to geometry, we're gonna be using a lot of algebra, but today the focus is more on equations and not on, for example, geometry. Okay, so let's start with the first questions. There were a certain number of people in a room. The average age of these people was equal to the number of people. Then a 29-year-old person came into the room and it turned out that the average age of the people in the room was still equal to the number of people in the room. How many people were there in the room at the beginning? Wait, that's not right. Average age, which is the mean, equal to the number of people. And the same average, still equal to the number of people which already increased by one. what variables you can use. Let me write the equations. I'm sure you probably at the last step of solving it. Some people already got the answer. So we start with let's call the number of people in the room right number of people in the room. At the beginning. And right, all these are our variables, then what we'd have we have the average age equal to the number of people so we have n equal to the sum of the age divided by n right and then another person come into the room, then we would have n plus one. Now, the number increased by one, and the sum of their age increases by 29 divided by n plus one. So I'm sure most of you wrote these two equations and set them up correctly. And now the question is how to find n right. So that is what we talked about last time is still absorbing equations in general, remote remote to my end over here we have some equal to n square, and then some last 29 equal to n plus one square. So now we don't know what the sum is and I just subtract the first equation from the second one to get rid of that some so have a 29 equal to n plus one square minus n square. So at this time, at this point, you can either guess right, you can plug in a few values of meaning in this case, the given so you can plug in n equal to 14 1516 and then find the answer. Or another way is last time we talked about identities, you remember the identities that we talked about last time we have if we have a and b, any two general numbers and a square minus b square equal to a plus b times a minus b. Right, so that is a very common entity identities that we use. So we come back over here I can factorize this so 29 would be equal to two n plus one n times one right because n plus one times one so we got it right away and equal to 14. So I think most of you probably maybe got stuck at the last step here. But as I said before, you can use, you know, guess and check just plug in to solve it in the last step. Go to the next one. So this question involving in both fighting missing numbers in a diagram, a three by three square initially has a number zero in each of its cells in one step, all four numbers in one two by two sub squares such as a shaded one. Yeah, shaded one. And then increased by one. This operation is repeated several times to obtain the arrangement on the right. Unfortunately, some numbers in this arrangement are hidden. What number is in the square with the question mark. So you probably need to quickly draw this diagram. So we need to unpack the question right we have a sub square here, two by two, every time I can choose any. How many, we have a four sub squares right another sub squares here. So we need to unpack the question right we have a sub square here, two by two, every time I can choose any. How many, we have a four sub squares right another sub squares here. We have four of them. So it doesn't matter which order we choose them. We can pick any of these sub squares. And increase all the numbers inside that sub square by one. and subscribe here. We asked for this number over here. So in principle, we have one, two, three, four, five, six unknowns, right? The question is that, do we want to assign each one variable for each of these unknowns? Or we want to do it differently, right? So that is the, you know, usually the crucial steps in solving algebra questions. Oh, great. A lot of you already figured out. I think you had a quick way to assign variables. So let's do it together. Yeah, you can certainly put the variables here to these unknowns. Whereas last time, you look at the question carefully and read the question carefully. It might not be the wisest way to choose to assign variables for each of these unknowns. But the way it works that every time we apply one operation and number in each sub-square increased by one. So we would figure out that maybe I want to assign variables as to the number of operation we apply to each sub-square. So I would from left to right, this is top left. I could say that maybe I did m operations and then top right n operations and bottom left. Maybe p and bottom right. That's the number of operations I do for each sub-squares. Then we see right away 18 is equal to m plus n. That's the total number of operation in the top two sub-squares. We have 18 equal to m plus n. We write all the equations that we can write. Then 13 here actually is equals to p right away. So 13 equal to p and 47, it gets contribution from all four sub-squares. So 47 equal to m plus n plus p, that's k. We are asked to find the k, the question mark which is equal to k. So we can see it right away after we write out these four equations that k is going to just equal to 47 minus 18 minus 13, m, n, p, k minus m, n, and p, which is 16. So majority got the correct answer. Let's go to the next one. This could be a very long questions, but I think David had an idea of how to do it in a nice way. So David, could you please help with this question? The positive integers a, b, and c have three digits each. For each integer, the first digit is the same as its last digit, which means it's a palindrome. Also, b equals 2a plus 1 and c equals 2b plus 1. How many possibilities are there for the integer a? So you should think a bit about what this means and play around with the numbers a bit because we find that we don't have to do algebra on a thousand different unknowns, right? We have very few and we can use that to help us cut down on our options a lot. It says, useful hint for bounding these numbers a little bit. Remember that they're all three-digit numbers. So for example, if a was 909, obviously b would not be three digits, which means that a has to be kind of small. And we can use that. So with that in mind, we'll have this is one of the last problems. We'll give it 45 more seconds, and then we'll talk about it a little bit. Okay, we're going to start. This is a hard problem. So again, if you have any ideas as we go along, make sure to use them, you know, in principle, you might have a bit more time to gone really fast to the rest of the test. So here's the idea. We know that A is some PQP and B is RSR CSTUT, right? If A has some hundreds and 10th digit P, right? Then B is going to be so that means that A is 100 P plus 10 Q plus P, right? And then B equals 200 P plus 20 Q plus two people's one. So what that means is that the hundredth digit of B is to P, right? Oh, so David, did you write a somehow I could not see your writing. Oh, I'm sorry. Okay, that's my okay. Sorry. There it is. Okay, let me actually just do it like this. So if A has, you know, PQP, this is just the one hard concept that one has to think of. B will have two P times 100, right? And then two Q times 10. And then two P plus one, right? So we want the hundredth digit to equal to the 10th digit, right? But it kind of looks like the hundredth digit is to P and the 10, the ones digit is two P plus one, right? That's a problem. So if we had 101, we'd get 203, right? These aren't equal, we want this to be three. So the way to do that is to realize that to Q must carry it must carry a one into the hundredths place so that, you know, it's a palindrome again, so that means that means that say A is PQP, that implies that Q is greater than or equal to five, right? Because we want it to carry into P. I hope that makes sense. If that doesn't happen, then obviously the hundredth digit of B will be less. But then we get this we have B has whatever the new units digit of Q is in the middle. And B has to follow the same property. I see you have a late joiner, all right, which means that B, when you multiply this by two, and then add one, the two Q must carry again. Again, to make sure that the hundredth digit of C matches the ones digit. So what this tells us is that Q carries over twice. So for example, if Q equals five, then five times two ends in zero, right, which doesn't carry again. And if Q equals six, six times two equals two, which doesn't carry again, you just get four, right? Two mod 10 really, which means that in fact, Q is eight or Q is nine. Because if Q is eight, then 16 will carry over to 32. Again, we're carrying over digits, which is what we have to do. So you've got P8P or P9P, right? Now I just want to find P. And what we realize here is that there's a very simple way of bounding this. I did talk about this before as a little hint, which is that say that A has digits two and two over here, then C will have digits five and five, sorry, B will have digits five and five and C will have four digits because 505 times two is 1000, which means that P has to be one, just because otherwise the numbers get really big really fast. So A is either 181 or A is 191. And we can check these. And we find that they both work. So the answer is actually two, which nobody put two, this is a very hard problem. And I can tell the more explain it, how tricky some of the concepts are. So let me just review really quick, the main two ideas. The main two ideas are, we have PQP for A, and we bound P to be one because otherwise A gets too big, that's very easy to check. And we bound Q to be greater than or equal to eight, so that when you multiply by four, you carry over to the 10th place two times. And that alone is enough to close this in. And interestingly, the algebra that we do is mostly just looking at the digits, right, saying that two Q has to carry over twice into 2P. But really, there's a lot of just looking at the numbers and looking at what the digits look like that goes into solving this problem. So it's conceptually difficult, but it's not really calculation difficult. If you have any more questions, let me know, because I realize this is a very tough problem. But that's basically the explanation. And that's basically how you get about to solving. Thank you, David. I think it's really nice. So mostly we are using inequalities. And I think bounding people to one, many people might be able to do that. But like this stuff about in Q bigger than eight is a hard, hard, hard step. So I would suggest that if this one is too hard, we can do start with A equal to MAM, right, PQP, as David said, PQP. And then right away, you can, I mean, after a few steps, you can bound people to one. So you could have A equal to 1Q1. And then from there, you can see that B is going to be equal to maybe 3R3, and C equal to 7S7. But that might be easier for most people. And from here, you can maybe do guess and check with Q rather than bounding this. I think it is very nice, but it could be hard for most people. So if you can make these steps, 1Q1, 3R3, and 7S7, just go over Q from zero to nine, you can quickly go over it and realize that there are only two choices. So the answer is 181 and 191, right? Those are the two answers. I agree. It's probably an easier competition solution. Okay, thank you. In the end, I think we had one student got correct answer to why you were explaining it. So that's great. Next one. Okay, this is a typical question with percent. I think we come back to our roots. So I just put one question in this topic. On a certain island, frogs are always either green or blue. The number of blue frogs increases by 60%, while the number of green frogs decreases by 60%. It turns out that the ratio of blue frogs to green frog is the same as the previous ratio in the opposite order, green frogs to blue frogs. By what percentage did the overall number of frogs change? So blue increases by 60%, the green, sorry, decreased by 60%. And the ratio now is like in opposite order, sort of flip the fraction. Oh, some people got either answer, but, you know, the ratio, the sort of flip, for example, initially you start with blue. So this is what it means by the ratio flip. So initially, for example, blue over green is one over, maybe two over five, for example. So after the change become green over blue is five over two, maybe. Might not be the correct ratio, but this is what it means. This is old ratio, and the new, this is like new green divides by new blue. It's gonna, the ratio has to be flipped. So A is not the correct answer. In this question, the choice of variables are, it's pretty straightforward, right? You can denote G as the number of green frogs in the beginning and B as the number of blue frogs in the beginning. And you can write the blue, the new blue and the new greens in terms of the old ones. Thank you, David. Yes, the new ratio is like the flip of the old ratio. So follow that line, we just see what we can do here. And this question, I just want to emphasize that we talk about percent increase and decrease. Is that one? To be careful about the formula. So we have B and B increases by 60%, right? So it's gonna become, the new B is gonna be 1.6 of the old B, that's what it means by percent. And by expressing the new one in terms of the old one, we don't have to make another variables. And G, because G decreases by 60%, so the new one's gonna be 0.4 B, right? You can just take B equal to 100. So 60%, 160, and G is 100 decreased by 60%. So it's gonna be 40, so that's 0.4 G. And now we have the ratio of the new one to the old one. We have a 1.6, I probably want to put that over here. 1.6 B over 0.4 G, right? This is a ratio of new blue to new green. It's the opposite of the old ratio, so it's gonna be G over B. So this is the most important equation in these questions. And from here, we just cross multiply these two fractions. We would have 1.6 B squared equal to 0.4 G squared, right? And from here, we divide it by 0.4, so we'd have a 4 B squared equal to G squared, or 2 B equals to G. And now we're asked to find by what percentage is the overall number of frogs change. So the change over here is 0.6 B, right? And the change over here is minus 0.6 G, right? So total change is equal to 0.6 B minus 0.6 G. And since we know that now we know that G equal to 2 B, so we can express everything in terms of B, so that would be easier, right? So this is gonna be 1.2 B minus 0.6, 1.2 G, no. Did I make any mistake over here? So G, yes, because it goes down, right? So 0.6 B is 0.6 B and minus 0.6 G is 1.2 G. So the overall change is decreased by minus 0.6 B. And the total we have is 2 B plus G, which is, and total in the beginning is 2 B plus G, which is 4 B. Did I make mistake somewhere, David? So finding that, yeah, 2 B equals G, that's right. So then you do, okay, so you start with, okay, so if we say that G is 2 B, you start with 3 B originally. Yes, so- And you end with 1.6 B plus 0.4 G, which is 1.6 B plus 0.8 B, because we plug in G is 2 B, that's 2.4 B. So now we have the difference between 2.4 B and 3 B, and that's just- Oh, I think it is correct because, so initially we have a 3 B in the beginning, right? And the total change is 0.6 B minus 0.6 G. So it's actually just minus 0.6 B. Exactly, so you go from 3 B, you take away 0.6 B, which is 20%, because 0.6 is a fifth of three. Yes, okay, I thought it's 4 B, so I made mistake in the last step here. So- See, algebra is hard, you know? 0.6- There are a lot of numbers, we all make mistakes sometimes. Yeah, and minus 3 B, so that's the minus 20%, right? But we don't care about increase or decrease, so the answer is 20%. Yeah, it's easy to make mistake and... Okay, good, actually, yeah, we have the majority of the people almost got the correct answers, so that's great. Go to the next questions about velocity. So David, your question. All right, so this is one of the long-anticipated physics questions. And by physics, we mean velocity, okay? No masses, don't freak out. A bicyclist rode the distance of 84 kilometers at a constant speed. Each hour, he rode two kilometers further than he planned, which shortened his whole ride by one hour. What was the bicyclist's speed? So again, just set up the relevant equations and solve. This is another five-point problem, so it's tricky, but it's also rather brute forcible. So if you don't find a nice trick, just solve the equation and it will work, guarantee. And we'll come back in a few minutes. Yeah, just a reminder of some basic equation. Distance is velocity times time. So David gave the hint of the choices variable as expected speed and T's expected time, you mean the planned one, right? The one that he initially planned, right? Yeah, let me fix that. That's a bit ambiguous, sorry. And remember, once you set up the equations and you couldn't quite solve it, you can just guess and check, go through the choices and check. Very good for multiple choice exams. That's very good, that's very good advice. Yeah. The whole point is to set up the equation and then the last step. Thank you. I'm going to go ahead and do that. So hopefully at this point we've reached the stop planning, start bashing stage. The equations are in here. I'll give a bit more time for just going through the computations because as of yet, we only have one answer in the poll. All right so as the answers begin to rack in let's begin going through the computations together shall we? We know that the plan so if we choose these variables where s is the planned speed and t is the span the planned time we know that s times t equals 84 because you know that's the distance distance to speed times time and we know that s plus 2 the faster speed times t minus 1 one hour less because it's shortened the whole ride by one hour is also 84 so now we need to solve for s so what we can do is we can expand the second equation we can go st plus 2t um yeah st plus 2t minus s minus 2 equals 84 which means okay it's kind of hard to write these but there we go if we expand that equation which means that um 2t equals s plus 2 right we're just subtracting the first equation from the second we also know something else this is another equation that I should have probably written down oh actually no okay that's not true but remember that t equals 84 over s right and we want to solve for s so that means that 2t is just 168 so now we have 2t is just 168 over s so we've got 2 times 84 over s equals s plus 2 that's um this equation over here and now we multiply both sides by s we get 168 equals s squared plus 2s and you know you can complete the square which is what I'm going to do I'm going to add one on both sides so this becomes 169 which means that we take this whole equation running out of room here we take the square root we get that s plus 1 equals 13 which means that the answer is not 12 but it's s plus 2 because again we want the actual speed which is 2 more than we planned so the answer is s plus 2 which is just 14 kilometers per hour and that would be d and everyone who answered which is um a good you know about them yeah yeah some majority of us has put 14 so we did the equations very well good job and yeah this one we know this one was very bashable took time of course but it is a five point problem I will say really quickly though as a little interlude that when I saw this problem first I just went if you know your 10 times tables sorry your 12 times tables you probably have seen something like 14 times 6 is 24 and 12 times 7 is 84 I know these by heart and so when I look at this problem I just go oh it's 14 you know no algebra needed which is just kind of a message that's saying it's useful to have the little arithmetic basics that we haven't really used much in years because that really shortens this problem to just you look at it and bam you know the answer but still we can do it the algebra way and everyone did the algebra right so excellent job I'm very proud of you guys and yeah yeah thank you David yes this is great yeah so I mean we end up with a quadratic equation sounds fancy but we can get away with that in like computing the square or just guess and check okay another basic questions train g which is traveling at a constant speed passes a milestone in eight seconds before meeting train h which is also traveling at a constant speed so two trains pass each other in nine seconds then train it passes the milestone in 12 seconds which of the following statement about the length of the train is true you know it looks like we have to solve for it because looking at these you know really don't know how we're gonna have it solve the problem sometimes the choice helps or sometimes you know they are just there so there's a little bit to unpack about these questions right train g they suppose it go in this direction it passes a milestone right something like a post and then it missed train h and then train it also go through the milestone after meeting train g that means it has to go in the opposite direction right so this is when g miss h right this is when a diagram is helpful and then after a while h also passes through this milestone the post over here it goes these directions so for g it takes eight seconds to pass the milestone and for h it takes 12 seconds right and when they are passing each other it takes 12 seconds of nine seconds for them to pass each other and the question asks what we can say about the relative length of the two trains So it takes Trin G only 8 seconds and Trin H 12 seconds. And what does it say? It could say that G is shorter than H or say G traveling at higher speed, right? So how do we translate that into some sort of equations? And because this 8 seconds means the time it takes for Trin G to travel a distance equal to its length. And one thing we notice at the beginning of the class is when the two objects traveling in opposite directions, their relative speed is the sum of their two speeds. So you can use that information to set up one equation here and two equations separately for G and H when they pass the milestone. Okay, interesting. We have A and D, which is opposite of algebra. I'm sure you know how to do that. It's just the algebra is kind of mixed up. So let's start by choosing variables. It's hard to know what's the best choice here, so I probably just do the conventional one. For each trend, I would need two variables, one for the length and one for the velocity. So I would have lg, simple g for length of g, and I call it maybe v1. And h is length of h, and then the velocity is v2. If you don't know how to make it simpler than that, then we just start with all the variables necessary to represent the physical situation here. So for g, we know that the length, the time equals to, okay, so distance is length of the trend g equal to velocity times time, right? So it's equal to 8v1. And then same for h, h equals 12v2. So that's two equations I can set up. And then when they're passing each other, right, then you can imagine like g, if you talk about frame of reference, it starts from here, the tip of the trend g here, and once it passes by completely, it goes to here. So the time, the distance traveled is actually g plus h. And the relative velocity is actually v1 plus v2, the sum of the two speeds. And then the time it takes is 9, right? So we have these three sets, a set of three equations, and we need to figure out the ratio of g over h. So from here, I just use substitution. I don't know where it leads me to, but I don't want v1 and v2 here, so I just try to eliminate them. So v1 is g divided by 8, and v2 is h divided by 12. So I substitute into here, I have 9, and this is g over 8, and v2 is h over 12. But now I know I'm good, right, because there's only g and h, so I can throw h on one side and g on the other side. So h here is 1, 9 over 12. If I bring it to the left-hand side, I have 3 over 12 h, and I bring g to the right-hand side, it's 9 over 8 minus 1, so it's 1 over 8 g, right? And then from here, I take 8 over here, so that's 24 divided by 12, so that's 2, so I have a 2h equals g. And that means g is twice as long as h, which is the first, actually the first person who got it got its correct answer. So I think the answer dh twice h, probably you just like flip the algebra, but you got the idea. So this is one of the more difficult questions in which we have to really know a little bit of physics to set up the equations correctly. What do you think, David? Yeah, I think, I mean, the idea of adding the velocities when they come together, all of it, it's one of those problems where this is a paradox that you often notice with hard math kangaroo problems and problems in general is every single step is obvious and intuitive, and duh, I could do that. But when you put it together, you get a problem that is really hard to wrap your head around. And that's just really interesting to me. Yeah, it's always, it can always be kind of frustrating when that happens, we have to remember that you have the ability to solve it, it's just going to require perseverance and a lot of mental effort. But if you can always do it, because you don't need a lot of complex physics knowledge to solve this problem, you just have to keep going. Yeah, the math is not that complex here, but really, you have to put small pieces together. I always find that one hour is really short. So there's another question, because if you sign up for Math Kangaroo, you know that you can get all the video solutions for free, right? And another question we prepared but didn't have time to go through, question 2013, question 28. We chose velocity because they just appear a lot in Math Kangaroo, and I think these are very interesting questions for physics, for algebra, so by all means, if you have time, please look at this question as well. Okay, so good. Thank you, everyone. Thank you, David, and we'll see you next week.

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